Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $y = \dfrac{9q + 27}{q + 10} \times \dfrac{-q + 2}{q^2 + q - 6} $
Answer: First factor the quadratic. $y = \dfrac{9q + 27}{q + 10} \times \dfrac{-q + 2}{(q + 3)(q - 2)} $ Then factor out any other terms. $y = \dfrac{9(q + 3)}{q + 10} \times \dfrac{-(q - 2)}{(q + 3)(q - 2)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ 9(q + 3) \times -(q - 2) } { (q + 10) \times (q + 3)(q - 2) } $ $y = \dfrac{ -9(q + 3)(q - 2)}{ (q + 10)(q + 3)(q - 2)} $ Notice that $(q - 2)$ and $(q + 3)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ -9\cancel{(q + 3)}(q - 2)}{ (q + 10)\cancel{(q + 3)}(q - 2)} $ We are dividing by $q + 3$ , so $q + 3 \neq 0$ Therefore, $q \neq -3$ $y = \dfrac{ -9\cancel{(q + 3)}\cancel{(q - 2)}}{ (q + 10)\cancel{(q + 3)}\cancel{(q - 2)}} $ We are dividing by $q - 2$ , so $q - 2 \neq 0$ Therefore, $q \neq 2$ $y = \dfrac{-9}{q + 10} ; \space q \neq -3 ; \space q \neq 2 $